Question

In a spring balance, the distance on the scale between the marking for the minimum mass (zero kilogram) and that for the maximum mass(5 kg) is 6 cm. When certain body is suspended from the hook of this balance the needle moves down by 1.2 cm. then the mass of the body is ____ kg.

• A. 3
• B. 2
• C. 4
• D. 1

Answer 1

Answer: (D)

1

$\begin{array}{cc}\text{max extension}& =6cm\\ x& =6\times {10}^{-2}m\\ \text{max extension}\text{at mass}=& 5kg\\ F=W=mg& =5\times 10\\ & =50N\end{array}$

$\begin{array}{c}F=kx\\ k=F/x=\frac{50}{6\times {10}^2}=\frac{5000}{6}\\ \text{for another body,}\\ \begin{array}{c}x=1.2cm=1.2\times {10}^{-2}m\\ \because F=kx\Rightarrow mg=kx\Rightarrow m=\frac{kx}{g}\end{array}\\ \Rightarrow m=\frac{5000}{6}\times \frac{1.2\times {10}^2}{10}=1kg\end{array}$